## Help with Laplace Transformations and 2nd order ODEs

so split it up:

$$\frac{1}{(s + 2)^3} + \frac{4}{(s + 2)^2}$$

$$\frac{4}{(s + 2)^2} = 4te^{-2t}$$

okay so for:

$$\frac{1}{(s + 2)^3}$$

we need to use:

$$t^n = \frac{n!}{s^{n + 1}}$$

so, would this mean:

$$\frac{3!}{s^3} = t^2$$

we have

$$\frac{1}{s^3}$$

Is this relevant, or am I going the wrong way for this one/

TFM

 for 1/(s+2)^3: laplace of t^2 is 2!/(s^3) so it is 2/s^3 i dont know why you put 1/s^3. but you have the right idea. so if the laplace of t^2 is 2!/(s^3), what is the inverse of 1/(s+2)^3 ?
 So: $$t^2 = \frac{2!}{s^3}$$ we have $$\frac{1}{(s + 2)^2}$$ so using the Shift, would that be something like: $$\frac{1}{(s + 2)^2} = (t - 2)^2$$ bvut I think even it is slightly wrong at least because it doesn't deal with the fact that the top value isn't 2! TFM
 For 1/(s+1)^3, let's just look at 1/s^3 ; we know that t^2 is 2/s^3 so 1/s^3 will be the inverse of (1/2)(t^2) right? but, we dont want 1/s^3, we want 1/(s+1)^3, so what do we do (like we did before)?

First Shft Theorem

 Ok, the first shift theorem says that: if you have a function that looks like: e^(at)F(t); the laplace of that is the laplace of F(t) but instead of putting p, you put p-a eg: L(e^3t(t)) we know that laplace of t is 1/p^2 (you know that from the tables right?) and so here a=3 so the laplace of e^(3t)(t)=1/(p-3)^2 so, where ever you have p, you replace with p-a in this case, p-3 Do you agree??
$$t^2 = 2/s^3$$

$$\frac{1}{2}t^2 = \frac{1}{s^3}$$

using FST

$$\frac{1}{2}t^2 = \frac{1}{b^3}$$, where b = s + 1

would this make it:

\frac{1}{2}(t+1)^2 = \frac{1}{b^3} [/tex]

???

TFM

 I dont see any e's in your answer (there should be). now as we did before, the second fraction is the laplace transform of: 4te^(-2t) because if you find the laplace of 4t, this will give you 4/s^2 but we dont want s at the bottom, we want (s+2) so thats why we use the first shift theorem and times the 4t by e^(-2t). now, can you do the same for the 1/(s+1)^3

Okay so:

 4t will give you 4/s^2,but we don't want s at the bottom, we want (s+2) so thats why we use the first shift theorem and times the 4t by e^(-2t).
so would this mean:

$$t^2 would give: \frac{2}{s^3}$$

thus would:

$$\frac{2}{(s + 1)^3} be \frac{1}{2}t^2 e^{-t}$$

???

TFM

 very good!!
 Great So now: $$Y = \frac{1}{(s + 2)^3} + \frac{4}{(s + 2)^2}$$ this goes to: $$Y = \frac{1}{2}t^2 e^{-t} + 4te^{-2t}$$ So now we put this value of Y back into: $$s^2(Y) - 4 + 4(sY) + 4Y = \frac{1}{s + 2}$$ ??? TFM
 y(t)=(1/2)t^2(e^-t) + 4t(e^-2t) This is the answer. dont substitute anything into anything.
 Excellent. So $$y(t) = \frac{1}{2}t^2 e^{-t} + 4te^{-2t}$$ Tomorrow, I'll start (c), if thats okay? TFM
 yep thats fine
 Okay, so: (c): $$y'' + y = sin(t), y_0 = 1, y_0' = 0$$ L(y(t)) = Y L(y'(t)) = sL(y(t)) - y(0) L(y''(t)) = sL(y'(t))-y'(0) = s^2(L(y(t))-sy(0)-y'(0) there is no y', $$L(y(t)) = Y$$ $$L(y'(t)) = sL(y(t)) - 1$$ $$L(y''(t)) = sL(y'(t))-0 = s^2(L(y(t))-sy(0)-y'(0)$$ and for the sin(t) $$sin(\alpha t) = \frac{\alpha}{p^2 + \alpha^2}$$ since alpha = 1: $$sin(t) = \frac{1}{p^2 + 1}$$ Is this okay, I am not quite sure because the lack of y' TFM
 That is absolutely right. what next?
 Okay so: $$L(y(t)) = Y$$ $$L(y'(t)) = sL(y(t)) - 1$$ $$L(y''(t)) = s^2(L(y(t))-sy(0)$$ So now I substitute these into the original equation: $$y'' + y = sin(t)$$ $$s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1}$$ Is this okay? Do I substitute y(0) into sy(0), or is there another value/leave it alone? TFM
 That's fine substitute y(0) and also substitue the value L(y(t))
 Okay, so $$s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1}$$ $$s^2(Y)-1 + Y = \frac{1}{p^2 + 1}$$ $$Ys^2 + Y - 1 = \frac{1}{p^2 + 1}$$ Now I need to make the Y the subject: $$Ys^2 + Y = \frac{1}{p^2 + 1} + 1$$ factorise out $$Y(s^2 + 1) = \frac{1}{p^2 + 1} + 1$$ divide through: $$Y = \frac{\frac{1}{p^2 + 1} + 1}{s^2 + 1}$$ Now I need to find the inverse. Firstly, split it up into two fractions: $$Y = \frac{\frac{1}{p^2 + 1}}{s^2 + 1} + \frac{1}{s^2 + 1}$$ Does this look okay? TFM